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Problems i just can't conquer.

  1. Apr 17, 2008 #1
    1) Find the Indefinite Integral.


    I have the first part: 1t-sin (3t^2)
    I'm not really sure about the (3t^2) part, then after that i'm lost.

    2)A vet has 100ft of fencing to construct 6 dog kennels by building a fence around a rectangular region, and then subdividing into 6 smaller rectangles by placing 5 fences parallel to on of the sides. What dimensions of the region will maximize the total area?

    So i made a rectangle, split it up into 6 squares labeling each vertical section "y", and each horizontal section "x"
    the perimeter equation would be P=13x+7y. Would the Area be 100ft since thats all he has to work with?

    Thanks in advance.
    Last edited: Apr 17, 2008
  2. jcsd
  3. Apr 17, 2008 #2
    For the first part, are you trying to integrate just the following, cuz i am a lill bit confused:

    [tex]\int\frac{1-cos(6t)}{2}dt[/tex] right??

    If so then this is just a tabelar integral since you can rewrite it in the form

    [tex]\int\frac{1-cos(6t)}{2}dt=\frac{1}{2}\int dt-\frac{1}{2}\int cos(6t)dt=.....????[/tex]

    Can you do this part now? if this was what you were asking at first place.
  4. Apr 17, 2008 #3


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    See sutupidmath's explanation for your first problem. (I have no idea where [itex]3t^{2}[/itex] came from).

    As for your second problem:
    100ft of fencing is not area that will be fenced in, but the physical fence. Meaning that your total perimeter will be 100. so one of your equations will be 100=sum(sides). there are 6 divisions on one side and 7 perpendicular divisions, therefore 12x+7y=100. You want to maximize the area so you need an equation for the area: A=side*side or A=y*6x (I chose y as the divisions and x as the sides). Do you see what to do from here?
  5. Apr 17, 2008 #4
    I think he tried to use the power rule for integrals in the function 'inside' the cos function.
  6. Apr 19, 2008 #5
    Would the first one be [tex]\frac{1}{2}x[/tex]_[tex]\frac{sin(6t)}{6}[/tex]?

    I have in my notes that [tex]\int cos(kx)dx= \frac{sin(kx)}{k}[/tex]+C

    For the second i have y=[tex]\frac{100-12x}{7}[/tex]

    so the area would A=[tex]\frac{600-72x^{2}}{7}[/tex]

    Would i take the derivative of the Area, set the top and bottom to 0 to find the critical points and thats how i find the maximum area?
  7. Apr 19, 2008 #6
    NOt really! Just be careful to add a constant C, and also make sure to factor 1/2 out.
  8. Apr 19, 2008 #7
    Factor out the 1/2?
  9. Apr 20, 2008 #8

    Gib Z

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    Homework Helper

    Did you forget about the factor of 1/2 outside the integral?
  10. Apr 20, 2008 #9
    What would i do with it? just subtract it from the sin(6t)/t?
  11. Apr 20, 2008 #10
    [tex]\int\frac{1-cos(6t)}{2}dt=\frac{1}{2}\int dt-\frac{1}{2}\int cos(6t)dt=\frac{1}{2}\left(t-\frac{1}{6}sin(6t)\right) +C[/tex]
  12. Apr 20, 2008 #11


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    First of all you need to review your arithmetic.

    Second, look at the equation for the area now. It's an upside down parabola. For what value of x would the maximum area occur?

    Review your notes on optimization if you still don't see it.
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